3.14.63 \(\int \frac {(1-3 k^2) x+2 k^2 x^3}{\sqrt [4]{(1-x^2) (1-k^2 x^2)} (-1+d+(-d+3 k^2) x^2-3 k^4 x^4+k^6 x^6)} \, dx\)

Optimal. Leaf size=98 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{k^2 x^2-1}\right )}{d^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{k^2 x^2-1}\right )}{d^{3/4}} \]

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Rubi [F]  time = 10.24, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (1-3 k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (-d+3 k^2\right ) x^2-3 k^4 x^4+k^6 x^6\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((1 - 3*k^2)*x + 2*k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/4)*(-1 + d + (-d + 3*k^2)*x^2 - 3*k^4*x^4 + k^6*
x^6)),x]

[Out]

(-4*k^2*(1 - x^2)^(1/4)*(1 - k^2*x^2)^(1/4)*Defer[Subst][Defer[Int][x^2/((1 - k^2 + k^2*x^4)^(1/4)*(-1 + d*x^4
- 3*k^2*(-1 + x^4) - 3*k^4*(-1 + x^4)^2 - k^6*(-1 + x^4)^3)), x], x, (1 - x^2)^(1/4)])/((1 - x^2)*(1 - k^2*x^
2))^(1/4) + (2*(1 - 3*k^2)*(1 - x^2)^(1/4)*(1 - k^2*x^2)^(1/4)*Defer[Subst][Defer[Int][x^2/((1 - k^2 + k^2*x^4
)^(1/4)*(1 - d*x^4 + 3*k^2*(-1 + x^4) + 3*k^4*(-1 + x^4)^2 + k^6*(-1 + x^4)^3)), x], x, (1 - x^2)^(1/4)])/((1
- x^2)*(1 - k^2*x^2))^(1/4) - (4*k^2*(1 - x^2)^(1/4)*(1 - k^2*x^2)^(1/4)*Defer[Subst][Defer[Int][x^6/((1 - k^2
+ k^2*x^4)^(1/4)*(1 - d*x^4 + 3*k^2*(-1 + x^4) + 3*k^4*(-1 + x^4)^2 + k^6*(-1 + x^4)^3)), x], x, (1 - x^2)^(1
/4)])/((1 - x^2)*(1 - k^2*x^2))^(1/4)

Rubi steps

\begin {align*} \int \frac {\left (1-3 k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (-d+3 k^2\right ) x^2-3 k^4 x^4+k^6 x^6\right )} \, dx &=\int \frac {x \left (1-3 k^2+2 k^2 x^2\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (-d+3 k^2\right ) x^2-3 k^4 x^4+k^6 x^6\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1-3 k^2+2 k^2 x}{\sqrt [4]{(1-x) \left (1-k^2 x\right )} \left (-1+d+\left (-d+3 k^2\right ) x-3 k^4 x^2+k^6 x^3\right )} \, dx,x,x^2\right )\\ &=\frac {\left (\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1-3 k^2+2 k^2 x}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (-1+d+\left (-d+3 k^2\right ) x-3 k^4 x^2+k^6 x^3\right )} \, dx,x,x^2\right )}{2 \sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {-1+3 k^2-2 k^2 x}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (1-d+\left (d-3 k^2\right ) x+3 k^4 x^2-k^6 x^3\right )} \, dx,x,x^2\right )}{2 \sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {3 \left (1-\frac {1}{3 k^2}\right ) k^2}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (1-d+\left (d-3 k^2\right ) x+3 k^4 x^2-k^6 x^3\right )}+\frac {2 k^2 x}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (-1+d-\left (d-3 k^2\right ) x-3 k^4 x^2+k^6 x^3\right )}\right ) \, dx,x,x^2\right )}{2 \sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (-1+d-\left (d-3 k^2\right ) x-3 k^4 x^2+k^6 x^3\right )} \, dx,x,x^2\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\left (-1+3 k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (1-d+\left (d-3 k^2\right ) x+3 k^4 x^2-k^6 x^3\right )} \, dx,x,x^2\right )}{2 \sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-1+x^4\right )}{\sqrt [4]{1+k^2 \left (-1+x^4\right )} \left (1-d x^4+3 k^2 \left (-1+x^4\right )+3 k^4 \left (-1+x^4\right )^2+k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (2 \left (-1+3 k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1+k^2 \left (-1+x^4\right )} \left (1-d x^4+3 k^2 \left (-1+x^4\right )+3 k^4 \left (-1+x^4\right )^2+k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-1+x^4\right )}{\sqrt [4]{1-k^2+k^2 x^4} \left (1-d x^4+3 k^2 \left (-1+x^4\right )+3 k^4 \left (-1+x^4\right )^2+k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (2 \left (-1+3 k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (1-d x^4+3 k^2 \left (-1+x^4\right )+3 k^4 \left (-1+x^4\right )^2+k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (-1+k^2 \left (3-3 k^2+k^4\right )+d \left (1-\frac {3 k^2 \left (-1+k^2\right )^2}{d}\right ) x^4-3 k^4 \left (1-k^2\right ) x^8-k^6 x^{12}\right )}+\frac {x^6}{\sqrt [4]{1-k^2+k^2 x^4} \left (1-k^2 \left (3-3 k^2+k^4\right )-d \left (1-\frac {3 k^2 \left (-1+k^2\right )^2}{d}\right ) x^4+3 k^4 \left (1-k^2\right ) x^8+k^6 x^{12}\right )}\right ) \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (2 \left (-1+3 k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (1-d x^4+3 k^2 \left (-1+x^4\right )+3 k^4 \left (-1+x^4\right )^2+k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (-1+k^2 \left (3-3 k^2+k^4\right )+d \left (1-\frac {3 k^2 \left (-1+k^2\right )^2}{d}\right ) x^4-3 k^4 \left (1-k^2\right ) x^8-k^6 x^{12}\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt [4]{1-k^2+k^2 x^4} \left (1-k^2 \left (3-3 k^2+k^4\right )-d \left (1-\frac {3 k^2 \left (-1+k^2\right )^2}{d}\right ) x^4+3 k^4 \left (1-k^2\right ) x^8+k^6 x^{12}\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (2 \left (-1+3 k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (1-d x^4+3 k^2 \left (-1+x^4\right )+3 k^4 \left (-1+x^4\right )^2+k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (-1+d x^4-3 k^2 \left (-1+x^4\right )-3 k^4 \left (-1+x^4\right )^2-k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt [4]{1-k^2+k^2 x^4} \left (1-d x^4+3 k^2 \left (-1+x^4\right )+3 k^4 \left (-1+x^4\right )^2+k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (2 \left (-1+3 k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (1-d x^4+3 k^2 \left (-1+x^4\right )+3 k^4 \left (-1+x^4\right )^2+k^6 \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}

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Mathematica [F]  time = 1.95, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1-3 k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (-d+3 k^2\right ) x^2-3 k^4 x^4+k^6 x^6\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((1 - 3*k^2)*x + 2*k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/4)*(-1 + d + (-d + 3*k^2)*x^2 - 3*k^4*x^4
+ k^6*x^6)),x]

[Out]

Integrate[((1 - 3*k^2)*x + 2*k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/4)*(-1 + d + (-d + 3*k^2)*x^2 - 3*k^4*x^4
+ k^6*x^6)), x]

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IntegrateAlgebraic [A]  time = 12.11, size = 98, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{-1+k^2 x^2}\right )}{d^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{-1+k^2 x^2}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 - 3*k^2)*x + 2*k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/4)*(-1 + d + (-d + 3*k^2)*x^2 - 3
*k^4*x^4 + k^6*x^6)),x]

[Out]

ArcTan[(d^(1/4)*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(1/4))/(-1 + k^2*x^2)]/d^(3/4) - ArcTanh[(d^(1/4)*(1 + (-1 - k^
2)*x^2 + k^2*x^4)^(1/4))/(-1 + k^2*x^2)]/d^(3/4)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*k^2+1)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(3*k^2-d)*x^2-3*k^4*x^4+k^6*x^6),x, alg
orithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, k^{2} x^{3} - {\left (3 \, k^{2} - 1\right )} x}{{\left (k^{6} x^{6} - 3 \, k^{4} x^{4} + {\left (3 \, k^{2} - d\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*k^2+1)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(3*k^2-d)*x^2-3*k^4*x^4+k^6*x^6),x, alg
orithm="giac")

[Out]

integrate((2*k^2*x^3 - (3*k^2 - 1)*x)/((k^6*x^6 - 3*k^4*x^4 + (3*k^2 - d)*x^2 + d - 1)*((k^2*x^2 - 1)*(x^2 - 1
))^(1/4)), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (-3 k^{2}+1\right ) x +2 k^{2} x^{3}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {1}{4}} \left (-1+d +\left (3 k^{2}-d \right ) x^{2}-3 k^{4} x^{4}+k^{6} x^{6}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*k^2+1)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(3*k^2-d)*x^2-3*k^4*x^4+k^6*x^6),x)

[Out]

int(((-3*k^2+1)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(3*k^2-d)*x^2-3*k^4*x^4+k^6*x^6),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, k^{2} x^{3} - {\left (3 \, k^{2} - 1\right )} x}{{\left (k^{6} x^{6} - 3 \, k^{4} x^{4} + {\left (3 \, k^{2} - d\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*k^2+1)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(3*k^2-d)*x^2-3*k^4*x^4+k^6*x^6),x, alg
orithm="maxima")

[Out]

integrate((2*k^2*x^3 - (3*k^2 - 1)*x)/((k^6*x^6 - 3*k^4*x^4 + (3*k^2 - d)*x^2 + d - 1)*((k^2*x^2 - 1)*(x^2 - 1
))^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {2\,k^2\,x^3-x\,\left (3\,k^2-1\right )}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/4}\,\left (3\,k^4\,x^4-d-k^6\,x^6+x^2\,\left (d-3\,k^2\right )+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*k^2*x^3 - x*(3*k^2 - 1))/(((x^2 - 1)*(k^2*x^2 - 1))^(1/4)*(3*k^4*x^4 - d - k^6*x^6 + x^2*(d - 3*k^2) +
1)),x)

[Out]

int(-(2*k^2*x^3 - x*(3*k^2 - 1))/(((x^2 - 1)*(k^2*x^2 - 1))^(1/4)*(3*k^4*x^4 - d - k^6*x^6 + x^2*(d - 3*k^2) +
1)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*k**2+1)*x+2*k**2*x**3)/((-x**2+1)*(-k**2*x**2+1))**(1/4)/(-1+d+(3*k**2-d)*x**2-3*k**4*x**4+k**6
*x**6),x)

[Out]

Timed out

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